D - Pair of Topics

Posted 2021-03-05 vetsama

The next lecture in a high school requires two topics to be discussed. The ii-th topic is interesting by aiaiunits for the teacher and by bibi units for the students.

The pair of topics ii and jj (i<ji<j) is called good if ai+aj>bi+bjai+aj>bi+bj (i.e. it is more interesting for the teacher).

Your task is to find the number of good pairs of topics.

Input

The first line of the input contains one integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the number of topics.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109), where aiai is the interestingness of the ii-th topic for the teacher.

The third line of the input contains nn integers b1,b2,…,bnb1,b2,…,bn (1≤bi≤1091≤bi≤109), where bibi is the interestingness of the ii-th topic for the students.

Output

Print one integer — the number of good pairs of topic.

Examples

Input

5
4 8 2 6 2
4 5 4 1 3

Output

7

Input

4
1 3 2 4
1 3 2 4

Output

0

这个题需要一点思路，ai+aj>bi+bj可以转换成ai-bi+aj-bj>0，也就是c[i]=a[i]-b[i]，只需要找c[i]+c[j]大于0

一开始的想法是枚举i和j，但是很显然会超时

a和b数组内部的顺序不会影响正确答案个数，所以可以排序

从两头缩进，如果a[r]加上比较小的a[l]大于0，那么就有l-r种方法，因为比a[l]大的数加上a[r]也会大于0，这样就能保证方法没有遗漏

当a[l]+a[r]不大于0的时候，说明a[l]太小了，l向右移来找一个更大的a[l]

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <algorithm>
using namespace std;



int main()
{
    int t;
    scanf("%d", &t);
    int a[200001] = { 0 };
    for (int i = 0; i < t; i++) {
        scanf("%d", &a[i]);
    }
    for (int i = 0; i < t; i++) {
        int k;
        scanf("%d", &k);
        a[i] -= k;
    }
    sort(a, a + t);
    int l = 0;
    int r = t - 1;
    long long sum = 0;


    while (l < r && a[r]>0) {
        if (a[l] + a[r] > 0) {
            if (l >= r)break;
            sum += (r - l);
            r--;
        }
        else {
            l++;
        }
    }
    cout << sum << endl;

    return 0;
}